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1963 IMO Problems/Problem 5

Revision as of 15:24, 8 December 2022 by Mathboy100 (talk | contribs)

Problem

Prove that $\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}$.

Solution

Because the sum of the $x$-coordinates of the seventh roots of unity is $0$, we have

\[\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} + \cos{\frac{8\pi}{7}} + \cos{\frac{10\pi}{7}} + \cos{\frac{12\pi}{7}} + \cos{\frac{14\pi}{7}} = 0\] \[\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} + \cos{\frac{8\pi}{7}} + \cos{\frac{10\pi}{7}} + \cos{\frac{12\pi}{7}} = -1\] \[\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} + \cos{-\frac{6\pi}{7}} + \cos{-\frac{4\pi}{7}} + \cos{-\frac{2\pi}{7}} = -1.\]

Now, we can apply $\cos{x} = \cos{-x}$ to obtain

\[2(\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}}) = -1\] \[\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} = -\frac{1}{2}.\]

Finally, since $\cos{x} = -cos{\pi-x}$,

\[\cos{\frac{2\pi}{7}} - \cos{\frac{3\pi}{7}} - \cos{\frac{\pi}{7}} = -\frac{1}{2}\] \[\cos{\frac{\pi}{7}} - \cos{\frac{2\pi}{7}} + \cos{\frac{3\pi}{7}} = \frac{1}{2}\textrm{. }\square\]

~mathboy100

Solution 2

Let $\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=S$. We have

\[S=\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\cos{\frac{\pi}{7}}+\cos{\frac{3\pi}{7}}+\cos{\frac{5\pi}{7}}\]

Then, by product-sum formulae, we have

\[S * 2* \sin{\frac{\pi}{7}} = \sin{\frac{2\pi}{7}}+\sin{\frac{4\pi}{7}}-\sin{\frac{2\pi}{7}}+\sin{\frac{6\pi}{7}}-\sin{\frac{4\pi}{7}}=\sin{\frac{6\pi}{7}}=\sin{\frac{\pi}{7}}\]

Thus $S = 1/2$. $\blacksquare$

Solution 3

Let $a=\sin{\frac{\pi}{7}}$ and $b=\cos{\frac{\pi}{7}}$. From the addition formulae, we have

\[S=b-[b^2-a^2]+[ b(b^2-a^2)-a(2ab) ]=b-b^2+b^3+a^2(1-3b)\]

From the Trigonometric Identity, $a^2=1-b^2$, so

\[S=b-b^2+b^3+(1-b^2)(1-3b)=4b^3-2b^2-2b+1\]

We must prove that $S=1/2$. It suffices to show that $8b^3-4b^2-4b+1=0$.

Now note that $\cos{\frac{4\pi}{7}}=-\cos{\frac{3\pi}{7}}$. We can find these in terms of $a$ and $b$:

\[\cos{\frac{4\pi}{7}}=[b^2-a^2]^2-[2ab]^2=b^4-6a^2b^2+b^4=b^4-6(1-b^2)b^2+b^4=8b^4-8b^2+1\]

\[\cos{\frac{3\pi}{7}}=b[b^2-a^2]-a[2ab]=b^3-3a^2b=b^3-3(1-b^2)b=3b-4b^3\]

Therefore $8b^4-8b^2+1=-(3b-4b^3)\Rightarrow 8b^4+4b^3-8b^2-3b+1=0$. Note that this can be factored:

\[8b^4+4b^3-8b^2-3b+1=(8b^3-4b^2-4b+1)(b+1)=0\]

Clearly $b\neq -1$, so $8b^3-4b^2-4b+1=0$. This proves the result. $\blacksquare$

Solution 4

Let $\omega=\mathrm{cis}\left(\frac{\pi}{14}\right)$. Thus it suffices to show that $\omega+\omega^{-1}-\omega^2-\omega^{-2}+\omega^3+\omega^{-3}=1$. Now using the fact that $\omega^k=\omega^{14+k}$ and $-\omega^2=\omega^9$, this is equivalent to \[\omega+\omega^3+\omega^5+\omega^7+\omega^9+\omega^{11}+\omega^{13}-\omega^7\] \[\omega\left(\frac{\omega^{14}-1}{\omega^2-1}\right)-\omega^7\] But since $\omega$ is a $14$th root of unity, $\omega^{14}=1$. The answer is then $-\omega^{7}=1$, as desired.

~yofro

See Also

1963 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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